x, y = 2.5, 2.5

$x_{1}, y_{1}$ = 2, 2

$x_{2}, y_{2}$ = 4, 2

$x_{3}, y_{4}$ = 4, 3

$x_{4}, y_{4}$ = 2, 4

Converting into-local coordinate,

Now, $\phi_{1}=\frac{1}{4}(1-\xi)(1-\eta)$

$\phi_{2} =\frac{1}{4} (1 + \xi)(1 − η)$

$\phi_{3} =\frac{1}{4} (1 + \xi)(1 − η)$

$\phi_{4} =\frac{1}{4} (1 + \xi)(1 − η)$

Now, $\bar{x} = \sum\phi_{j}\bar{x}_{j}$

∴ $\bar{x}$ = $\phi_{1}\bar{x}_{1}+\phi_{1}\bar{x}_{ 2} +\phi_{1}\bar{x}_{3}+\phi_{1}\bar{x}_{ 4}$

0.5 = $\frac{1}{4}(1 − ξ)(1 − η)\bar{x}_{1} + \frac{1}{4}(1 + ξ)(1 − η)\bar{x}_{2} + \frac{1}{4} (1 + ξ)(1 + η)\bar{x}_{3} + \frac{1}{4}(1 − ξ)(1 + η)\bar{x}_{1}$

i.e. $\frac{1}{2}=\frac{1}{4}(1+\xi)(1-\eta)\ast 2+\frac{1}{4}(1-\xi)(1+\eta)\ast 2$

i.e $\frac{1}{2}= \frac{1}{2}(1 + ξ)(1 − η) +\frac{1}{2} + (1 + ξ)(1 + η)$

1=$1+\xi+\eta-\eta\xi+1+\xi+\eta+\eta\xi$

1=2+2$\xi$

$\xi \frac{-1}{2}$

= (1 + ξ)(1 − η) + (1 + ξ)(1 + η)

Now, $\bar{y}=\sum \phi_{j}\bar{y}_{j}$

0.5=$\frac{1}{4}0.5 = (1 − ξ)(1 − η) \bar{y}_{1}+ \frac{1}{4}(1 + ξ)(1 − η)bar{x}_{2} + \frac{1}{4}(1 + ξ)(1 + η)\bar{x}_{3} + \frac{1}{4}(1 − ξ)(1 + η)\bar{x}_{4}$

0.5 =$\frac{1}{4} (1 + ξ)(1 + η) + \frac{1}{4}(1 − ξ)(1 + η)$

$\frac{1}{2}=\frac{1}{4}[1+\xi+\eta+\eta\xi+1-\xi+\eta-\eta\xi$

∴ 2 = 2 +$ 2η =\gt 2η = 0 =\gt η = 0$

Therefore, point P is given by (-1/2,0) in $\xi, -\eta$coordinate,

To compute the result,

$\phi_{1} = \frac{1}{4}(1 − ξ)(1 − η) = \frac{1}{4}∗ (1 + \frac{1}{2})∗ 1 =\frac{3}{8}$

$\phi_{2} = \frac{1}{4}(1 − ξ)(1 − η) = \frac{1}{4}∗ (1 + \frac{1}{2})∗ 1 =\frac{1}{8}$

$\phi_{3} = \frac{1}{4}(1 − ξ)(1 − η) = \frac{1}{4}∗ (1 + \frac{1}{2})∗ 1 =\frac{1}{8}$

$\phi_{4} = \frac{1}{4}(1 − ξ)(1 − η) = \frac{1}{4}∗ (1 + \frac{1}{2})∗ 1 =\frac{3}{8}$

Now,

$\phi_{1} + \phi_{2}+\phi_{3}+\phi_{4}=\frac{3}{8}+\frac{1}{8}+\frac{1}{8}+\frac{3}{8}=\frac{8}{8}$

$\phi_{1} + \phi_{2}+\phi_{3}+\phi_{4}$

Now, Temperature at point P,

$T_P = \phi_{1}T_{1} + \phi_{2}T_{2} + \phi_{3}T_{3} + \phi_{4}T_{4}$

∴ $T_{1} = 100^{\circ} C T_{2} = 60^{\circ}C T_{3} =50^{\circ} C T_{4} = 90^{\circ}C$

∴ $T_P =\frac{3}{8} ∗ 100 + \frac{1}{8}∗ 60 +\frac{1}{8} ∗ 50 +\frac{3}{8} ∗ 90$

$T_{P}=85^{\circ}C$